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Erdős ProblemsCombinatorics

Erdős Problem 282: Graham

Graham has shown that mn\frac{m}{n} is the sum of distinct unit fractions with denominators ≡a(modd)\equiv a\pmod{d} if and only if (n(n,a,d),d(a,d))=1.\left(\frac{n}{(n,a,d)},\frac{d}{(a,d)}\right)=1. Does the greedy algorithm always terminate in such cases?

CombinatoricsAMS 5

Mathematical statement

Graham has shown that mn\frac{m}{n} is the sum of distinct unit fractions with denominators ≡a(modd)\equiv a\pmod{d} if and only if (n(n,a,d),d(a,d))=1.\left(\frac{n}{(n,a,d)},\frac{d}{(a,d)}\right)=1. Does the greedy algorithm always terminate in such cases?

Statement source: Erdős Problems statement material

Statement terms: Source-specific

Source-specific terms. Therefore does not assert reuse rights beyond attributed display.

Statement artifacts, not proofs

These records expose exact Lean propositions and statement-only wrappers. Defining a proposition does not supply a proof of it. A placeholder-bearing target also contains no proof. Elaboration checks syntax and types; it does not certify that a formalization perfectly captures every nuance of the informal problem.

Pinned Lean formulation 1

erdos_282.variants.graham

Canonical source
Complete statement target, proof intentionally absentLean 4
theorem erdos_282.variants.graham {x : ℚ} (hx : x ∈ Set.Ioo 0 1) {a d : ℕ} (hd : 1 < d)    (h : (x.den / x.den.gcd (a.gcd d)).gcd (d / a.gcd d) = 1) :    (greedyUnitFractionRem { n | n ≡ a [MOD d] } x =ᶠ[atTop] 0) ↔ answer(sorry) := by  sorry
Statement source
Formal Conjectures
Lean version
v4.27.0
Placeholder
Present; no proof artifact
Source evidence
Pinned source index
Fidelity review
Community formulation

References

Pinned Formal Conjectures sourceformalizationerdosproblems.com/282background